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Increasing GP with AMs 6 and 54: Find AM of a, d | JEE

JEE Maths question with a full step-by-step solution.

Question
a,b,c,da, b, c, d are in increasing G.P. If the A.M. between aa and bb is 66 and the A.M. between cc and dd is 5454, then the A.M. of aa and dd is
A1515
B4848
C4444
D4242correct
Solution
Step 1: Let the ratio be rr, so b=ar, c=ar2, d=ar3b=ar,\ c=ar^2,\ d=ar^3. The two given A.M.s become
a+ar2=6 ⇒ a(1+r)=12,ar2+ar32=54 ⇒ ar2(1+r)=108.\frac{a+ar}{2}=6\ \Rightarrow\ a(1+r)=12,\qquad \frac{ar^2+ar^3}{2}=54\ \Rightarrow\ ar^2(1+r)=108.
 Step 2: Divide the second by the first:
ar2(1+r)a(1+r)=10812 ⇒ r2=9 ⇒ r=3(r=−3 rejected, G.P. is increasing).\frac{ar^2(1+r)}{a(1+r)}=\frac{108}{12}\ \Rightarrow\ r^2=9\ \Rightarrow\ r=3\quad(r=-3\text{ rejected, G.P. is increasing}).
 Step 3: Then a(1+3)=12⇒a=3a(1+3)=12\Rightarrow a=3, and d=ar3=3⋅27=81d=ar^3=3\cdot 27=81. Step 4:
a+d2=3+812=42.\frac{a+d}{2}=\frac{3+81}{2}=42.
 Correct answer: (4)
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